Question: 85. The standard heat of formation, in \mathrm{kcal} / \mathrm{mol} of \mathrm{Ba}^{2+} is: [Given : standard heat of formation of \mathrm{SO}_{4}^{2-} ion (\mathrm{aq})=-216 \mathrm{kcal} / \mathrm{mol}, standard heat of crystallization of \mathrm{BaSO}_{4}(\mathrm{~s})=-4.5 \mathrm{kcal} / \mathrm{mol}, standard heat of formation of \mathrm{BaSO}_{4}(\mathrm{~s})=-349 \mathrm{kcal} / \mathrm{mol} ]
(1) -128.5
(2) -133.0
(3) +133.0
(4) +220.5
Answer: Option (1)
Explanation:
The standard heat of crystallization of barium sulphate is defined for the reaction:
\mathrm{Ba}^{2+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq}) \rightarrow \mathrm{BaSO}_{4}(\mathrm{s}) \Delta H_{\text{crystallization}}=-4.5 \ \mathrm{kcal/mol}Using Hess’s law, the standard heat of formation of \mathrm{BaSO}_{4}(\mathrm{s}) is given by:
\Delta H_f(\mathrm{BaSO}_{4})=\Delta H_f(\mathrm{Ba}^{2+})+\Delta H_f(\mathrm{SO}_{4}^{2-})+\Delta H_{\text{crystallization}}Substituting the given values:
-349=\Delta H_f(\mathrm{Ba}^{2+})+(-216)+(-4.5)Simplifying:
-349=\Delta H_f(\mathrm{Ba}^{2+})-220.5 \Delta H_f(\mathrm{Ba}^{2+})=-349+220.5 \Delta H_f(\mathrm{Ba}^{2+})=-128.5 \ \mathrm{kcal/mol}Therefore, the correct answer is option (1).